Electric field around a circuit

  • #1
Est120
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TL;DR Summary
How the electric field varies around a circuit?
We know that E is conservative so the integral of E around a closed loop is zero. I know this helps us (in some way, that's why i'm asking) to calculate the total voltage drop around the complete circuit (which is zero).
What exactly is "E" in the integral? For example, internet says "electric field is zero inside a conductor" so, then what? E. dl = 0 and we are integrating zero around the closed loop. There's nothing interesting there.
Inside a battery for example there is a non-conservative E, which can't be used in the integral formula.
There must be an electric field inside the conductor in some way, something must push charges.

I would want to know an explanation (that makes sense) of int( E.dl) = 0 and how that relates to the change in voltage around the closed circuit.
 
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  • #2
Est120 said:
There must be an electric field inside the conductor in some way, something must push charges.
That is not the case.

Power does not flow in the wire, it flows in the electric and magnetic fields that surround the two conductors of a closed circuit.

The current flows on the surface of a good conductor. That current is "driven" by the magnetic field, as the magnetic filed is guided by the surface of the conductor.

The surface of the conductor is also an equipotential, that guides the electric field.
 
  • #3
That current is "driven" by the magnetic field, as the magnetic filed is guided by the surface of the conductor.
Magnetic field appears basically nowhere in these equations.
 

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  • #4
The diagram you show is for a resistor, not for a good conductor such as a wire.
Where did you find that diagram?

See: The Big Misconception About Electricity.
 
  • #5
Est120 said:
TL;DR Summary: How the electric field varies around a circuit?

We know that E is conservative so the integral of E around a closed loop is zero. I know this helps us (in some way, that's why i'm asking) to calculate the total voltage drop around the complete circuit (which is zero).
What exactly is "E" in the integral? For example, internet says "electric field is zero inside a conductor" so, then what? E. dl = 0 and we are integrating zero around the closed loop. There's nothing interesting there.
Inside a battery for example there is a non-conservative E, which can't be used in the integral formula.
There must be an electric field inside the conductor in some way, something must push charges.

I would want to know an explanation (that makes sense) of int( E.dl) = 0 and how that relates to the change in voltage around the closed circuit.
First, the electric field is not necessarily conservative, but let's forget about this and consider a simple circuit without (explicit) magnetic components. To simplify further, let's consider a real circuit formed by a battery, some copper wire and a resistor.

Since copper has a finite conductivity, there will be an electric field inside the wires. It will be small, but it will be non zero. Typical value for 20 mA in a 1.25 diameter copper wire is about 20 uV/m (iirc).
The field will be much stronger in the resistor, in accordance to Ohm's law E = j / sigma, for example tens or hundreds of volts per meter.

The electric field inside wires and resistors will follow the direction of the wires and the resistive material. How is it possible? Because of the surface and interface charge that develops during the initial very fast transient when you close the circuit.

I am not getting into the battery because the field in there will depend on the particular chemistry and construction, but know that the surface and interface charge develops because the battery creates a separation of charges that is associated with and electric field in the space around it.

Once you reach the steady state where E_mat = j / sigma_mat for the various materials (copper, carbon...) you can easily compute the line integrals knowing that the field is oriented like the portion of circuit where you evaluate it.
The line integral inside the battery will give you the EMF, the line integral inside the conductor will give you a minimal ohmic loss (usually approximated to zero) and the line integral inside the resistive material will give you the voltage drop 'along' the resistor.

Since we are considering the conservative case, the sum of all line integrals along the whole circuit (the circulation of E) will be zero. That is KVL.

For a detailed walkthrough for a circuit with lumped R L and C components see Ramo Whinnery VanDuzer, "Field and Waves in communication electronics", chapter 4, " The electromagnetics of circuits".
 
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  • #6
Baluncore said:
The current flows on the surface of a good conductor. That current is "driven" by the magnetic field, as the magnetic filed is guided by the surface of the conductor.
That is not correct imho. You are speaking of perfect or super-conductors, and this is not the case in the near totality of circuits. Copper is a good conductor and has finite conductivity. Current will flow throughout the entire section of the wire at DC and you need moderately high frequencies to see skin effect.

The assumption that wires in a circuit have zero resistance should be a simplification to reflect the zero-dimensionality of the circuit connections in the lumped model. Pretending we are using superconductors can only cause confusion, like that experienced by the op.
 
  • #7
SredniVashtar said:
Since copper has a finite conductivity, there will be an electric field inside the wires. It will be small, but it will be non zero. Typical value for 20 mA in a 1.25 diameter copper wire is about 20 uV/m (iirc).
20 uV/m is quite insignificant when you consider the typical voltages used in circuits. There are thermal junction voltages higher than that. The resistive voltage drop, is usually exceeded by the voltages induced by changes in load current.

SredniVashtar said:
The electric field inside wires and resistors will follow the direction of the wires and the resistive material. How is it possible?
That is as obvious as Ohms law.
SredniVashtar said:
Because of the surface and interface charge that develops during the initial very fast transient when you close the circuit.
I don't believe you.
When you connect the battery to the circuit, the initial transient current goes to charge the capacitance of the transmission line. (Every wire is a part of a greater transmission line). The applied voltage appears initially across the inductance of the wires, so the current to the load increases. That builds the magnetic field, that in turn maintains the current.

In a good conductor, the current is intimately connected with the magnetic field, not with the electric field.
 
  • #8
Baluncore said:
20 uV/m is quite insignificant when you consider the typical voltages used in circuits. There are thermal junction voltages higher than that. The resistive voltage drop, is usually exceeded by the voltages induced by changes in load current.

20 uV/m is the exact electric field required to sustain the full 20 mA current flowing in the circuit. It does not matter how small it is wrt to the other fields in the circuit. In the wires conductivity is 6.1 10^7 S/m, so 20 uV/m are a significant and non-negligible field.

That is as obvious as Ohms law.

And Ohm's law is upheld thanks to the field produced by surface charge that superposes to the field created by the battery to create the resultant field we see inside the components.

I don't believe you.

Would you believe John David Jackson?

When you connect the battery to the circuit, the initial transient current goes to charge the capacitance of the transmission line. (Every wire is a part of a greater transmission line).

I see you are focusing on the field outside the components, which is also produced by surface charge. Of course the charge set in motion inside the wires produces a magnetic field and, given the proper conditions, some circuits can be considered to be transmission lines. But we must be careful not to overstretch the model of a transmission line. A transmission line is... a line, and has such has a preferential direction of transmission (and a mathematical model that neglects propagation delays in the transversal direction). It is nonetheless based on the same phenomena of redistribution of surface and interface charge (and the ensuing variations in magnetic flux), but this is beyond the level of explanation required by the OP
 
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  • #9
SredniVashtar said:
It is nonetheless based on the same phenomena of redistribution of surface and interface charge (and the ensuing variations in magnetic flux), but this is beyond the level of explanation required by the OP
Which is why we can ignore the resistance of the copper wires that form the nodes of the DC circuit. If the resistance was significant, we would use thicker wire.
 
  • #10
Baluncore said:
Which is why we can ignore the resistance of the copper wires that form the nodes of the DC circuit. If the resistance was significant, we would use thicker wire.
The OP was asking what makes the charge in the wires move. The answer is: the electric field in the wire. It is small but not zero. The example I showed has a 20uV/m electric field that - as small as it is - is responsible for the 20 mA current in the 1.25 mm diameter copper wire.

A wire of this diameter has negligible resistance in every practical circuit at lab scale, and at the current give it will have negligible resistance. No need to use a thicker wire. Most importantly, charge will flow (at DC) through the whole cross section of the wire and not only on its surface as it would happen with a superconducting wire at near zero Kelvin temperature.

Pushing the idea that having practically zero resistance means the wires obey the quantum physics of superconductor is an educational error, in my opinion, because it transforms a simplification into a complication.
 
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  • #11
SredniVashtar said:
For a detailed walkthrough for a circuit with lumped R L and C components see Ramo Whinnery VanDuzer, "Field and Waves in communication electronics", chapter 4, " The electromagnetics of circuits".
Sorry, I was reading chapter 4 from the above book. He writes down the line integral ##\oint E \cdot dl## along the circuit closed path decomposing it in different terms. About the first term it is equal to the voltage of the source/generator (suppose a battery).

Since inside the battery take place chemical processes, then inside it there are actually two forces acting on the charges/electrons: the Coulomb force from the Electric field ##E## and the electromotive force say ##f##.

I believe he considers only the Electric field in the equation since inside the battery the electric field ##E=f##.
 
  • #12
cianfa72 said:
Sorry, I was reading chapter 4 from the above book. He writes down the line integral ##\oint E \cdot dl## along the circuit closed path decomposing it in different terms. About the first term it is equal to the voltage of the source/generator (suppose a battery).

Since inside the battery take place chemical processes, then inside it there are actually two forces acting on the charges/electrons: the Coulomb force from the Electric field ##E## and the electromotive force say ##f##.

I believe he considers only the Electric field in the equation since inside the battery the electric field ##E=f##.

Is there a question here? The electric field at the jump between battery terminals is equal to minus the work per unit charge done by nonelectric forces inside the battery. All other lines integrals are evaluated inside the components and when it comes to the inductors, it's just a piece of wire with (almost) zero field inside (the coil contribution comes from the other side of Faraday's equation).
 
  • #13
SredniVashtar said:
The electric field at the jump between battery terminals is equal to minus the work per unit charge done by nonelectric forces inside the battery.
Yes, the line integral of the electric field ##E## along a path between the battery terminals is exactly equal to the electromotive force of the battery (due to internal nonelectric chemical processes).
 
  • #14
Non-ideal conductors always tend to be in electrostatic equilibrium and that means the electric field always tends to be zero in the conductors.

When there is not a potential difference applied across a non-ideal conductor, the current through the conductor will be zero and the electric field in the conductor will be zero too. The conductor will be in electrostatic equilibrium.

When there is a potential difference applied across a non-ideal conductor the current through the conductor will not be zero and the electric field in the conductor will not be zero too. Although electrons move through the conductor trying to eliminate the presence of the electric field, the conductor will not be brought into electrostatic equilibrium because of the potential difference applied across the conductor which is not zero.

Est120 said:
TL;DR Summary: How the electric field varies around a circuit?

What exactly is "E" in the integral? For example, internet says "electric field is zero inside a conductor" so, then what? E. dl = 0 and we are integrating zero around the closed loop.
So we are not integrating zero around the closed loop.
 
  • #15
Gavran said:
So we are not integrating zero around the closed loop.
Note that since in DC the EM field is static then the integral of Electric field along the closed circuit loop (i.e. its circulation) is actually null. However if you look at the electromotive force inside the battery its integral over a path along the battery is not null.
 
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